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A 0.2-kg block on a horizontal, frictionless surface is attached to a horizontal spring. The spring...

A 0.2-kg block on a horizontal, frictionless surface is attached to a horizontal spring. The spring constant is k = 600 N/m. The block is pulled to the right until it is a distance of 0.08 m from the unstrained position and released from rest. What is the kinetic energy of the block when it is 0.06 m from the unstrained position?

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Answer #1

ma 0.2kg k= boon)m OL= 0.08m. change in Oc! = 0.06m The KE attained by the block = PE of the spring KE= 2 ko? 2 14(c) 2 x 6

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